我知道如何获得两个平面列表的交集：
b1 = [1,2,3,4,5,9,11,15]
b2 = [4,5,6,7,8]
b3 = [val for val in b1 if val in b2]
或者
def intersect(a, b):
return list(set(a) & set(b))
print intersect(b1, b2)
但是当我必须找到嵌套列表的交集时，我的问题就开始了：
c1 = [1, 6, 7, 10, 13, 28, 32, 41, 58, 63]
c2 = [[13, 17, 18, 21, 32], [7, 11, 13, 14, 28], [1, 5, 6, 8, 15, 16]]
最后我想收到：
c3 = [[13,32],[7,13,28],[1,6]]
你们能帮我解决这个问题吗？
您不需要定义交集。它已经是套装的一流部分。
>>> b1 = [1,2,3,4,5,9,11,15]
>>> b2 = [4,5,6,7,8]
>>> set(b1).intersection(b2)
set([4, 5])
如果你想：
c1 = [1, 6, 7, 10, 13, 28, 32, 41, 58, 63]
c2 = [[13, 17, 18, 21, 32], [7, 11, 13, 14, 28], [1, 5, 6, 8, 15, 16]]
c3 = [[13, 32], [7, 13, 28], [1,6]]
那么这里是 Python 2 的解决方案：
c3 = [filter(lambda x: x in c1, sublist) for sublist in c2]
在 Python 3filter中返回一个 iterable 而不是list，因此您需要使用 包装filter调用list()：
c3 = [list(filter(lambda x: x in c1, sublist)) for sublist in c2]
解释：
过滤器部分获取每个子列表的项目并检查它是否在源列表 c1 中。对 c2 中的每个子列表执行列表推导。
对于只想找到两个列表的交集的人，Asker 提供了两种方法：
b1 = [1,2,3,4,5,9,11,15] b2 = [4,5,6,7,8] b3 = [val for val in b1 if val in b2]和
def intersect(a, b): return list(set(a) & set(b)) print intersect(b1, b2)
但是有一种更有效的混合方法，因为您只需在列表/集合之间进行一次转换，而不是三个：
b1 = [1,2,3,4,5]
b2 = [3,4,5,6]
s2 = set(b2)
b3 = [val for val in b1 if val in s2]
这将在 O(n) 中运行，而他涉及列表理解的原始方法将在 O(n^2) 中运行
函数式方法：
input_list = [[1, 2, 3, 4, 5], [2, 3, 4, 5, 6], [3, 4, 5, 6, 7]]
result = reduce(set.intersection, map(set, input_list))
它可以应用于1+列表的更一般情况
>>> c1 = [1, 6, 7, 10, 13, 28, 32, 41, 58, 63]
>>> c2 = [[13, 17, 18, 21, 32], [7, 11, 13, 14, 28], [1, 5, 6, 8, 15, 16]]
>>> c1set = frozenset(c1)
展平变体：
>>> [n for lst in c2 for n in lst if n in c1set]
[13, 32, 7, 13, 28, 1, 6]
嵌套变体：
>>> [[n for n in lst if n in c1set] for lst in c2]
[[13, 32], [7, 13, 28], [1, 6]]
& 运算符取两个集合的交集。
{1, 2, 3} & {2, 3, 4}
Out[1]: {2, 3}
采用 2 个列表的交集的 Pythonic 方式是：
[x for x in list1 if x in list2]
您应该使用此代码进行展平（取自http://kogs-www.informatik.uni-hamburg.de/~meine/python_tricks），该代码未经测试，但我很确定它有效：
def flatten(x):
"""flatten(sequence) -> list
Returns a single, flat list which contains all elements retrieved
from the sequence and all recursively contained sub-sequences
(iterables).
Examples:
>>> [1, 2, [3,4], (5,6)]
[1, 2, [3, 4], (5, 6)]
>>> flatten([[[1,2,3], (42,None)], [4,5], [6], 7, MyVector(8,9,10)])
[1, 2, 3, 42, None, 4, 5, 6, 7, 8, 9, 10]"""
result = []
for el in x:
#if isinstance(el, (list, tuple)):
if hasattr(el, "__iter__") and not isinstance(el, basestring):
result.extend(flatten(el))
else:
result.append(el)
return result
展平列表后，以通常的方式执行交集：
c1 = [1, 6, 7, 10, 13, 28, 32, 41, 58, 63]
c2 = [[13, 17, 18, 21, 32], [7, 11, 13, 14, 28], [1, 5, 6, 8, 15, 16]]
def intersect(a, b):
return list(set(a) & set(b))
print intersect(flatten(c1), flatten(c2))
既然intersect定义了，一个基本的列表理解就足够了：
>>> c3 = [intersect(c1, i) for i in c2]
>>> c3
[[32, 13], [28, 13, 7], [1, 6]]
由于 S. Lott 的评论和 TM. 的相关评论，改进：
>>> c3 = [list(set(c1).intersection(i)) for i in c2]
>>> c3
[[32, 13], [28, 13, 7], [1, 6]]
鉴于：
> c1 = [1, 6, 7, 10, 13, 28, 32, 41, 58, 63]
> c2 = [[13, 17, 18, 21, 32], [7, 11, 13, 14, 28], [1, 5, 6, 8, 15, 16]]
我发现以下代码运行良好，如果使用 set 操作可能更简洁：
> c3 = [list(set(f)&set(c1)) for f in c2]
它得到：
> [[32, 13], [28, 13, 7], [1, 6]]
如果需要订购：
> c3 = [sorted(list(set(f)&set(c1))) for f in c2]
我们有：
> [[13, 32], [7, 13, 28], [1, 6]]
顺便说一句，对于更 Python 的风格，这个也很好：
> c3 = [ [i for i in set(f) if i in c1] for f in c2]
我不知道我是否迟到了回答你的问题。在阅读了您的问题后，我想出了一个可以在列表和嵌套列表上工作的函数 intersect()。我用递归来定义这个函数，很直观。希望它是您正在寻找的：
def intersect(a, b):
result=[]
for i in b:
if isinstance(i,list):
result.append(intersect(a,i))
else:
if i in a:
result.append(i)
return result
例子：
>>> c1 = [1, 6, 7, 10, 13, 28, 32, 41, 58, 63]
>>> c2 = [[13, 17, 18, 21, 32], [7, 11, 13, 14, 28], [1, 5, 6, 8, 15, 16]]
>>> print intersect(c1,c2)
[[13, 32], [7, 13, 28], [1, 6]]
>>> b1 = [1,2,3,4,5,9,11,15]
>>> b2 = [4,5,6,7,8]
>>> print intersect(b1,b2)
[4, 5]
您是否考虑[1,2]与 相交[1, [2]]？也就是说，你关心的只是数字，还是列表结构？
如果只有数字，请研究如何“扁平化”列表，然后使用该set()方法。
我也在寻找一种方法来做到这一点，最终结果是这样的：
def compareLists(a,b):
removed = [x for x in a if x not in b]
added = [x for x in b if x not in a]
overlap = [x for x in a if x in b]
return [removed,added,overlap]
c1 = [1, 6, 7, 10, 13, 28, 32, 41, 58, 63]
c2 = [[13, 17, 18, 21, 32], [7, 11, 13, 14, 28], [1, 5, 6, 8, 15, 16]]
c3 = [list(set(c2[i]).intersection(set(c1))) for i in xrange(len(c2))]
c3
->[[32, 13], [28, 13, 7], [1, 6]]
我们可以为此使用 set 方法：
c1 = [1, 6, 7, 10, 13, 28, 32, 41, 58, 63]
c2 = [[13, 17, 18, 21, 32], [7, 11, 13, 14, 28], [1, 5, 6, 8, 15, 16]]
result = []
for li in c2:
res = set(li) & set(c1)
result.append(list(res))
print result
要定义正确考虑元素基数的交集，请使用Counter：
from collections import Counter
>>> c1 = [1, 2, 2, 3, 4, 4, 4]
>>> c2 = [1, 2, 4, 4, 4, 4, 5]
>>> list((Counter(c1) & Counter(c2)).elements())
[1, 2, 4, 4, 4]
# Problem: Given c1 and c2:
c1 = [1, 6, 7, 10, 13, 28, 32, 41, 58, 63]
c2 = [[13, 17, 18, 21, 32], [7, 11, 13, 14, 28], [1, 5, 6, 8, 15, 16]]
# how do you get c3 to be [[13, 32], [7, 13, 28], [1, 6]] ?
这是一种c3不涉及集合的设置方法：
c3 = []
for sublist in c2:
c3.append([val for val in c1 if val in sublist])
但如果你喜欢只使用一行，你可以这样做：
c3 = [[val for val in c1 if val in sublist] for sublist in c2]
这是列表推导中的列表推导，这有点不寻常，但我认为你应该不会有太多麻烦。
c1 = [1, 6, 7, 10, 13, 28, 32, 41, 58, 63]
c2 = [[13, 17, 18, 21, 32], [7, 11, 13, 14, 28], [1, 5, 6, 8, 15, 16]]
c3 = [list(set(i) & set(c1)) for i in c2]
c3
[[32, 13], [28, 13, 7], [1, 6]]
对我来说，这是一种非常优雅和快速的方法:)
如果重复很重要，请使用此方法
from collections import Counter
def intersection(a, b):
"""
Find the intersection of two iterables
>>> intersection((1,2,3), (2,3,4))
(2, 3)
>>> intersection((1,2,3,3), (2,3,3,4))
(2, 3, 3)
>>> intersection((1,2,3,3), (2,3,4,4))
(2, 3)
>>> intersection((1,2,3,3), (2,3,4,4))
(2, 3)
"""
return tuple(n for n, count in (Counter(a) & Counter(b)).items() for _ in range(count))
def difference(a, b):
"""
Find the symmetric difference of two iterables
>>> difference((1,2,3), (2,3,4))
(1, 4)
>>> difference((1,2,3,3), (2,3,4))
(1, 3, 4)
>>> difference((1,2,3,3), (2,3,4,4))
(1, 3, 4, 4)
"""
diff = lambda x, y: tuple(n for n, count in (Counter(x) - Counter(y)).items() for _ in range(count))
return diff(a, b) + diff(b, a)
from random import *
a = sample(range(0, 1000), 100)
b = sample(range(0, 1000), 100)
print(a)
print(b)
print(set(a).intersection(set(b)))